Kondo Effect in a Fermionic Hierarchical Model

Abstract

In this paper, a fermionic hierarchical model is defined, inspired by the Kondo model, which describes a 1-dimensional lattice gas of spin-1/2 electrons interacting with a spin-1/2 impurity. This model is proved to be exactly solvable, and is shown to exhibit a Kondo effect, i.e. that, if the interaction between the impurity and the electrons is antiferromagnetic, then the magnetic susceptibility of the impurity is finite in the 0-temperature limit, whereas it diverges if the interaction is ferromagnetic. Such an effect is therefore inherently non-perturbative. This difficulty is overcome by using the exact solvability of the model, which follows both from its fermionic and hierarchical nature.

Appendices

Appendix 1: Comparison with the Original Kondo Model

If the partition function for the original Kondo model in presence of a magnetic field h acting only on the impurity site and at finite L is denoted by $Z^0_K(\beta ,\lambda _0,h)$ and the partition function for the model Eq. (2.1) with the same field h is denoted by $Z_K(\beta ,\lambda _0,h)$, then

$$\begin{aligned} Z_K(\beta ,\lambda _0,h)=Z^0_K(\beta ,\lambda _0,h)+Z^0_K(\beta ,0,0) \end{aligned}$$

(8.1)

so that by defining

$$\begin{aligned} \kappa \,\,{\buildrel def\over =}\,\,1+\frac{Z^0_K(\beta ,0,0)}{Z^0_K(\beta ,\lambda _0,h)} \end{aligned}$$

(8.2)

we get

$$\begin{aligned} m_K(\beta ,\lambda _0,h)= & {} \frac{1}{\kappa }{m^0_K(\beta ,\lambda _0,h)}, \nonumber \\ m^0_K(\beta ,\lambda _0,h)= & {} \kappa \, m_K(\beta ,\lambda _0,h) \\ \chi _K(\beta ,\lambda _0,h)= & {} \frac{1}{\kappa }\,{\chi _K^0(\beta ,\lambda _0,h)} +\frac{\kappa -1}{\kappa }\,\beta \,m^0_K(\beta ,\lambda _0,h)^2\nonumber \\ \chi _K^0(\beta ,\lambda _0,h)= & {} \kappa \, \chi _K(\beta ,\lambda _0,h)-(\kappa -1)\beta m_K(\beta ,\lambda _0,h)^2.\nonumber \end{aligned}$$

(8.3)

In addition $1\le \kappa \le 2$: indeed the first inequality is trivial and the second follows from the variational principle (see [16, Theorem 7.4.1, p. 188]):

$$\begin{aligned} \log Z^0_K(\beta ,\lambda _0,h)= & {} \max _\mu (s(\mu )-\mu (H_0+V))\nonumber \\\ge & {} s(\mu _0)-\mu _0(H_0)+\mu _0(V)=s(\mu _0)-\mu _0(H_0)\nonumber \\= & {} \log Z^0_K(\beta ,0,0) \end{aligned}$$

(8.4)

where $s(\mu )$ is the entropy of the state $\mu$, and in which we used

$$\begin{aligned} \mu _0(V)=\mathrm{Tr}\,( e^{-\beta H_0}\,V)/Z_K(\beta ,0,0)=0. \end{aligned}$$

(8.5)

Therefore, for $\beta h^2\ll 1$ (which implies that if there is a Kondo effect then $\beta m_K^2\ll 1$), the model Eq. (2.1) exhibits a Kondo effect if and only if the original Kondo model does, therefore, for the purposes of this paper, both models are equivalent.

Appendix 2: Some Identities

In this appendix, we state three relations used to compute the flow equation Eq. (5.13), which follow from a patient algebraic meditation:

$$\begin{aligned} {\langle \,A_1^{j_1}A_2^{j_2}\,\rangle }= & {} \delta _{j_1,j_2}\Big (2 +\frac{1}{3}\mathbf{a}^2\Big )-2\,a^{j_1,j_2}\delta _{j_1\ne j_2}\, s_{t_2,t_1} \nonumber \\ {\langle \,A^{j_1}_1A^{j_2}_1A^{j_3}_2\,\rangle }\equiv & {} 2\,a^{j_3}\,\delta _{j_1,j_2}\\ {\langle \,A_1^{j_1}A_1^{j_2}A_2^{j_3}A_2^{j_4}\,\rangle }= & {} 4\delta _{j_1,j_2}\delta _{j_3,j_4}\nonumber \end{aligned}$$

(9.1)

where the lower case $\mathbf{a}$ denote ${\langle \,\mathbf{A}_1\,\rangle }\equiv {\langle \,\mathbf{A}_2\,\rangle }$ and $a^{j_1,j_2}={\langle \,\psi ^+_1\sigma ^{j_1}\sigma ^{j_2}\psi ^-_1\,\rangle } ={\langle \,\psi ^+_2\sigma ^{j_1}\sigma ^{j_2}\psi ^-_2\,\rangle }$.

Appendix 3: Complete Beta Function

The beta function for the flow described in Sect. (6) is

$$\begin{aligned} \ell _0^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \ell _0 -2\ell _0\ell _6 +18\ell _0\ell _3 +3 \ell _0\ell _2 +3 \ell _0\ell _1 -2\ell _0^2\right) \nonumber \\ \ell _1^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \frac{1}{2}\ell _1 +9\ell _2\ell _3 +\frac{3}{2}\ell _8^2 +\frac{1}{12}\ell _6^2 +\frac{1}{2}\ell _5\ell _7 +\frac{1}{24}\ell _4^2 +\frac{1}{6}\ell _0\ell _6 +\frac{1}{4}\ell _0^2\right) \nonumber \\ \ell _2^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( 2\ell _2+36\ell _1\ell _3 + \ell _0^2 +6\ell _7^2 +\frac{1}{3}\ell _6^2 +\frac{1}{6}\ell _5^2 +2\ell _4\ell _8 +\frac{2}{3}\ell _0\ell _6\right) \\ \ell _3^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \frac{1}{2}\ell _3 +\frac{1}{4}\ell _1\ell _2 +\frac{1}{24} \ell _0^2 +\frac{1}{36}\ell _0\ell _6 +\frac{1}{72}\ell _6^2 +\frac{1}{12}\ell _5\ell _7 + \frac{1}{12}\ell _4\ell _8 \right) \nonumber \\ \ell _4^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \ell _4 +6\ell _6\ell _7 +\ell _5\ell _6 +108\ell _3\ell _8 +18\ell _2\ell _8 +3\ell _1\ell _4 +6\ell _0\ell _7 +\ell _0\ell _5 \right) \nonumber \\ \ell _5^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( 2\ell _5 +12 \ell _6\ell _8 +2\ell _4\ell _6 +216 \ell _3\ell _7 +6\ell _2\ell _5 +36\ell _1\ell _7 +12\ell _0\ell _8 +2\ell _0\ell _4\right) \nonumber \\ \ell _6^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \ell _6 +18\ell _7\ell _8 +3\ell _5\ell _8 +3\ell _4\ell _7 +\frac{1}{2}\ell _4\ell _5 +18\ell _3\ell _6 +3\ell _2\ell _6 +3\ell _1\ell _6 +2\ell _0\ell _6\right) \nonumber \\ \ell _7^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \frac{1}{2}\ell _7 +\frac{1}{2}\ell _6\ell _8 +\frac{1}{12}\ell _4\ell _6 +\frac{3}{2}\ell _3\ell _5 +\frac{3}{2}\ell _2\ell _7 +\frac{1}{4}\ell _1\ell _5 +\frac{1}{2}\ell _0\ell _8 +\frac{1}{12}\ell _0\ell _4\right) \nonumber \\ \ell _8^{[m-1]}= & {} \frac{1}{C^{[m]}}\left( \ell _8 +\ell _6\ell _7 +\frac{1}{6}\ell _5\ell _6 +3\ell _3\ell _4 +\frac{1}{2}\ell _2\ell _4 +3\ell _1\ell _8 +\ell _0\ell _7 +\frac{1}{6}\ell _0\ell _5\right) \nonumber \\ C^{[m]}= & {} 1+ 2\ell _0^2+(\ell _0+\ell _6)^2 +9\ell _1^2 +9\ell _2^2 +324\ell _3^2 +\frac{1}{2}\ell _4^2 +\frac{1}{2}\ell _5^2 +18\ell _7^2 +18\ell _8^2 \nonumber \end{aligned}$$

(10.1)

in which we dropped the $^{[m]}$ exponent on the right side. By considering the linearized flow equation (around $\ell _j = 0$), we find that $\ell _0,\ell _4,\ell _6,\ell _8$ are marginal, $\ell _2,\ell _5$ relevant and $\ell _1,\ell _3,\ell _7$ irrelevant. The consequent linear flow is very different from the full flow discussed in Sect. 6.

The vector ${\varvec{\ell }}$ is related to ${\varvec{\alpha }}$ via the following map:

$$\begin{aligned} \ell _0&=\alpha _0,\quad \ell _1=\alpha _1+\frac{1}{12}\alpha _4^2,\quad \ell _2=\alpha _2+\frac{1}{12}\alpha _5^2\nonumber \\ \ell _3&=\alpha _3+\frac{1}{12}\alpha _0^2+\frac{1}{18}\alpha _0\alpha _6+\frac{1}{2}\alpha _1\alpha _2 +\frac{1}{6}\alpha _4\alpha _8 +\frac{1}{6}\alpha _5\alpha _7+\frac{1}{36}\alpha _6^2 \nonumber \\&\quad +\,\frac{1}{36}\alpha _0\alpha _4\alpha _5 +\frac{1}{24}\alpha _1\alpha _5^2 +\frac{1}{24}\alpha _2\alpha _4^2 +\frac{1}{36}\alpha _4\alpha _5\alpha _6 +\frac{1}{288}\alpha _4^2\alpha _5^2 \\ \ell _4&=\alpha _4,\quad \ell _5=\alpha _5,\quad \ell _6=\alpha _6+\frac{1}{2}\alpha _4\alpha _5\nonumber \\ \ell _7&=\alpha _7+\frac{1}{6}\alpha _0\alpha _4+\frac{1}{2}\alpha _1\alpha _5+\frac{1}{6}\alpha _4\alpha _6 +\frac{1}{24}\alpha _4^2\alpha _5 \nonumber \\ \ell _8&=\alpha _8+\frac{1}{6}\alpha _0\alpha _5+\frac{1}{2}\alpha _2\alpha _4+\frac{1}{6}\alpha _5\alpha _6 +\frac{1}{24}\alpha _4\alpha _5^2.\nonumber \end{aligned}$$

(10.2)

Appendix 4: The Algebra of the Operators $O_{n,\pm }$

Lemma 1

Given $\eta \in \{-,+\}$, $m\le 0$ and $\Delta \in \mathcal Q_m$, the span of the operators $\{O_{n,\eta }^{[\le m]}(\Delta )\}_{n\in \{0,1,2,3\}}$ defined in Eq. (5.6) is an algebra, that is all linear combinations of products of $O_{n,\eta }^{[\le m]}(\Delta )$’s is itself a linear combination of $O_{n,\eta }^{[\le m]}(\Delta )$’s. The same result holds for the span of the operators $\{O_{n,\eta }^{[\le m]}(\Delta )\}_{n\in \{0,\ldots ,8\}}$ defined in Eq. (6.5).

Proof

The only non-trivial part of this proof is to show that the product of two $O_{n,\eta }$’s is a linear combination of $O_{n,\eta }$’s. $\square$

Due to the anti-commutation of Grassmann variables, any linear combination of $\psi _{\alpha }^{[\le m]\pm }$ and $\varphi _{\alpha }^{[\le m]\pm }$ squares to 0. Therefore, a straightforward computation shows that $\forall (i,j)\in \{1,2,3\}^2$,

$$\begin{aligned} A^{i}_\eta A^{j}_\eta =2\delta _{i,j} \psi ^{+}_{\uparrow } \psi ^{+}_{\downarrow } \psi ^{-}_{\uparrow } \psi ^{-}_{\downarrow },\quad B^{i}_\eta B^{j}_\eta =2\delta _{i,j} \varphi ^{+}_{\uparrow } \varphi ^{+}_{\downarrow } \varphi ^{-}_{\uparrow } \varphi ^{-}_{\downarrow } \end{aligned}$$

(11.1)

where the labels $^{[\le m]}$ and $(\Delta )$ are dropped to alleviate the notation. In particular, this implies that any product of three $A^{i}_\eta$ for $i\in \{1,2,3\}$ vanishes (because the product of the right side of the first of Eq. (11.1) and any Grassmann field $\psi _{\alpha }^{\pm }$ vanishes) and similarly for the product of three $B^{i}_\eta$.

Using Eq. (11.1), we prove that $\mathrm {span}\{O_{n,\eta }^{[\le m]}(\Delta )\}_{n\in \{0,1,2,3\}}$ is an algebra. For all $n\in \{0,1,2,3\}$, $p\in \{1,2,3\}$, $l\in \{1,2\}$,

$$\begin{aligned} O_{p}^2=0,\quad O_{3}O_{n}=0,\quad O_{l}O_{0}=0,\quad O_{0}^2=\frac{1}{6}O_{3},\quad O_{1}O_{2}=\frac{1}{2}O_{3} \end{aligned}$$

(11.2)

(here the $^{[\le m]}$, $(\Delta )$ and $_\eta$ are dropped). This concludes the proof of the first claim.

Next we prove that $\mathrm {span}\{O_{n,\eta }^{[\le m]}(\Delta )\}_{n\in \{0,\ldots ,8\}}$ is an algebra. In addition to Eq. (11.2), we have, for all $p\in \{0,\ldots ,8\}$,

$$\begin{aligned}&O_0O_4=\frac{1}{6}O_7,\quad O_0O_5=\frac{1}{6}O_8,\quad O_0O_6=\frac{1}{18}O_3,\quad O_0O_7=O_0O_8=0, \nonumber \\&O_1O_5=\frac{1}{2}O_7,\quad O_1O_4=O_1O_6=O_1O_7=O_1O_8=0,\quad O_2O_4=\frac{1}{2}O_8, \nonumber \\&O_2O_5=O_2O_6=O_2O_7=O_2O_8=0,\quad O_3O_p=0,\quad O_4^2=\frac{1}{6}O_1, \\&O_4O_5=\frac{1}{2}O_6,\quad O_4O_8=\frac{1}{6}O_3,\quad O_4O_7=0,\quad O_5^2=\frac{1}{6}O_2,\quad O_5O_7=\frac{1}{6}O_3, \nonumber \\&O_5O_8=0,\quad O_6^2=\frac{1}{18}O_3,\quad O_6O_7=O_6O_8=0,\quad O_7^2=O_8^2=O_7O_8=0. \nonumber \end{aligned}$$

(11.3)

This concludes the proof of the lemma.

Appendix 5: Fixed Points at $h=0$

We first compute the fixed points of Eq. (5.13) for $\ell _2\ge 0$. It follows from Eq. (5.13) that if ${\varvec{\ell }}$ is a fixed point, then $\ell _1=6\ell _3$, which implies

$$\begin{aligned} (1-3\ell _2)\left( \ell _2(1+3\ell _2)+6\ell _1^2+\ell _0^2 \right) =0. \end{aligned}$$

(12.1)

If $\ell _2\ge 0$, Eq. (12.1) implies that either $\ell _2=\ell _1=\ell _0=0$ or $\ell _2=\frac{1}{3}$. In the latter case, either $\ell _0=\ell _1=0$ or $\ell _0\not =0$ and Eq. (5.13) becomes

(12.2)

In particular, $\ell _1(1-12\ell _1)>0$, so that

$$\begin{aligned} \ell _0=\pm 2\sqrt{\frac{\ell _1(1+18\ell _1^2)}{1-12\ell _1}} \end{aligned}$$

(12.3)

which we inject into Eq. (12.2) to find that $\ell _0<0$ and

$$\begin{aligned} 1-\frac{35}{4}(3\ell _1)+\frac{27}{2}(3\ell _1)^2-\frac{19}{4}(3\ell _1)^3+107(3\ell _1)^4=0. \end{aligned}$$

(12.4)

Finally, we notice that $\frac{1}{12}$ is a solution of Eq. (12.4), which implies that

$$\begin{aligned} 4-19(3\ell _1)-22(3\ell _1)^2-107(3\ell _1)^3=0 \end{aligned}$$

(12.5)

which has a unique real solution. Finally, we find that if $\ell _1$ satisfies Eq. (12.5), then

$$\begin{aligned} 2\sqrt{\frac{\ell _1(1+18\ell _1^2)}{1-12\ell _1}}=3\ell _1\frac{1+15\ell _1}{1-12\ell _1}. \end{aligned}$$

(12.6)

We have therefore shown that, if $\ell _2\ge 0$, then Eq. (5.13) has three fixed points:

$$\begin{aligned}&{\varvec{\ell }}_0^*:=(0,0,0,0),\quad {\varvec{\ell }}_+^*:=\left( 0,0,\frac{1}{3},0\right) ,\nonumber \\&{\varvec{\ell }}^*:=\left( -x_0\frac{1+5x_0}{1-4x_0},\frac{x_0}{3},\frac{1}{3},\frac{x_0}{18}\right) . \end{aligned}$$

(12.7)

In addition, it follows from Eqs. (5.13) and (5.11) that, if $\lambda _0<0$, then (recall that $\alpha _0^{[0]}=\lambda _0$ and $\alpha _i^{[0]}=0$, $i=1,2,3$)

$$\begin{aligned} \ell _0^{[m]}<0,\quad 0\le \ell _2^{[m]}<\frac{1}{3},\quad 0\le \ell _1^{[m]}<6\ell _3^{[m]}<\frac{1}{12} \end{aligned}$$

(12.8)

for all $m\le 0$, which implies that the set $\{{\varvec{\ell }}\ |\ \ell _0<0,\ \ell _2\ge 0,\ \ell _1\ge 0,\ \ell _3\ge 0\}$ is stable under the flow. In addition, if $\ell _0^{[m]}>-\frac{2}{3}$, then $\ell _0^{[m-1]}<\ell _0^{[m]}$, so that the flow cannot converge to ${\varvec{\ell }}_0^*$ or ${\varvec{\ell }}_+^*$. Therefore if the flow converges, then it converges to ${\varvec{\ell }}^*$.

We now study the reduced flow Eq. (5.17), and prove that starting from $-2/3<\ell ^{[0]}_0<0$, $\ell ^{[0]}_2=0$, the flow converges to $f^*$. It follows from Eq. (5.17) that $\ell ^{[m]}_0<0$, $\ell ^{[m]}_2>0$ for all $m<0$, so that if Eq. (5.17) converges to a fixed point, then it must converge to $f^*$. In addition, by a straightforward induction, one finds that $\ell _2^{[m-1]}>\ell _2^{[m]}$ if $\ell _2^{[m]}<\frac{1}{3}$. Furthermore, $(2\ell _2^{[m]}+(\ell _0^{[m]})^2)\le \frac{1}{3}C^{[m]}$, which implies that $\ell _2^{[m]}\le \frac{1}{3}$. Therefore $\ell _2^{[m]}$ converges as $m\rightarrow -\infty$. In addition, $\ell _0^{[m-1]}<\ell _0^{[m]}$ if $\ell _0^{[m]}>-\frac{2}{3}$, and $\ell _0^{[m]}>-\frac{1}{3}-\ell _2^{[m]}\ge -\frac{2}{3}$, so that $\ell _0^{[m]}$ converges as well as $m\rightarrow -\infty$. The flow therefore tends to $f^*$.

Finally, we prove that starting from $\ell ^{[0]}_0>0,\ell ^{[0]}_2=0$, the flow converges to $f_+$. Similarly to the anti-ferromagnetic case, $\ell ^{[m]}_2>0$ for all $m<0$, $\ell _2^{[m]}\le \frac{1}{3}$ and $\ell _2^{[m-1]}>\ell _2^{[m]}$. In addition, by a simple induction, if $\lambda _0<1$, then $\ell _0^{[m]}>0$ and $\ell _0^{[m]}+\frac{1}{3}-\ell _2^{[m]}$ is strictly decreasing and positive. In conclusion, $\ell _0^{[m]}$ and $\ell _2^{[m]}$ converge to $f_+$.

Fig. 7

Plot of $n_j(\lambda _0)|\lambda _0|$ for $j=0$ (blue, color online) and $j=1,3$ (red) as a function of $|\log _{10}|\lambda _0||$. This plot confirms Eq. (6.10)

Fig. 8

Plot of $n_2(\lambda _0)|\log _2|\lambda _0||^{-1}$ as a function of $|\log _{10}|\lambda _0||$. This plot confirms Eq. (6.11)

Fig. 9

Plot of $r_j(h)|\log _2(h)||$ as a function of $|\log _{2}(h)|$. This plot confirms Eq. (6.12)

Appendix 6: Asymptotic Behavior of $n_j(\lambda _0)$ and $r_j(h)$

In this appendix, we show plots to support the claims on the asymptotic behavior of $n_j(\lambda _0)$ (see Eq. 6.10, Fig. 7 and Eq. 6.11, Fig. 8) and $r_j(h)$ (see Eq. 6.12, Fig. 9). The plots below have error bars which are due to the fact that $n_j(\lambda _0)$ and $r_j(h)$ are integers, so their value could be off by ${\pm }1$.

Appendix 7: Kondo Effect, XY-Model, Free Fermions

In [1], given $\nu \in [1,\ldots ,L]$, the Hamiltonian $H_h=H_0 {-h} \,\sigma _\nu ^z,$ with

$$\begin{aligned} H_0={- \frac{1}{4}} \sum _{n=1}^L \left( \sigma ^x_n\sigma ^x_{n+1}+\sigma ^y_n\sigma ^y_{n+1}\right) . \end{aligned}$$

(13.1)

has been considered with suitable boundary conditions, under which $H_0$ and ${\sigma ^z_0} +1$ are unitarily equivalent to $\sum _{q}{(-\cos q)} \, a^+_qa^-_q$ and, respectively, to $\frac{2}{L} \sum _{q,q'} a^+_q a^-_{q'} e^{i\nu (q-q')}$ in which $a^\pm _q$ are fermionic creation and annihilation operators and the sums run over q’s that are such that $e^{iq L}=-1$. It has been shown, [1]³, that, by defining

$$\begin{aligned} F_L(\zeta )= & {} 1+\frac{2 h}{L} \sum _q \frac{1}{\zeta +\cos q}\nonumber \\ F(z)= & {} \lim _{L\rightarrow \infty } F_L(z)=1+\frac{2\,h}{\pi }\,\int _0^\pi \frac{dq}{(z+\cos q)} \end{aligned}$$

(13.2)

the partition function is equal to $Z_L^0\zeta _L$ in which $Z_L^0$ is the partition function at $h=0$ and is extensive (i.e. of $O(e^{const L})$) and (see Appendix 8, Eq. 14.12)

$$\begin{aligned} \log {\zeta _L (\beta ,h)}=-\beta h +\frac{1}{2\pi i}\oint _C \log \left( 1+{e^{-\beta z}} \right) \Big [\frac{\partial _zF_L(z)}{F_L(z)} \Big ]\,dz \end{aligned}$$

(13.3)

where the contour C is a closed curve in the complex plane which contains the zeros of $F_L(\zeta )$ (e.g. , for $L\rightarrow \infty$, a curve around the real interval $[-1,\sqrt{1+4h^2}]$ if $h<0$ and $[-\sqrt{1+4h^2},1]$ if $h>0$) but not those of $1+e^{-\beta z}$ (which are on the imaginary axis and away from 0 by at least $\frac{\pi }{\beta }$). In addition, it follows from a straightforward computation that $(F(z)-1)/h$ is equal to the analytical continuation of $2 (z^2-1)^{-\frac{1}{2}}$ from $(1,\infty )$ to $C\setminus [-1,1]$.

At fixed $\beta <\infty$ the partition function $\zeta _L(\beta ,h)$ has a non extensive limit $\zeta (\beta ,h)$ as $L\rightarrow \infty$; $\zeta (\beta ,h)$ and the susceptibility and magnetization values $m(\beta ,h)$ and $\chi (\beta ,h)$, are given in the thermodynamic limit by

$$\begin{aligned} \log \zeta (\beta ,h)= & {} -\beta h {+\frac{\beta }{2\pi i}} \oint _C\frac{dz}{1+{e^{\beta z}}} \log \left( 1 {+} \frac{2h}{(z^2-1)^{\frac{1}{2}}}\right) \nonumber \\ m(\beta ,h)= & {} -1+\frac{1}{\pi i}\oint _C \frac{1}{1+{e^{\beta z}}} \frac{dz}{(z^2-1)^{\frac{1}{2}}{+2h} }\\ \chi (\beta ,h)= & {} -\frac{2}{\pi i}\oint _C \frac{1}{1+{e^{\beta z}}} \frac{dz}{((z^2-1)^{\frac{1}{2}}{+2h})^2}\nonumber \end{aligned}$$

(13.4)

so that $\chi (\beta ,0)=\frac{2\sinh (\beta )}{(1+\cosh (\beta ))}$ and, in the $\beta \rightarrow \infty$ limit,

$$\begin{aligned} m(\infty ,h)= \frac{2h}{\sqrt{1+4h^2}} ,\quad \chi (\infty ,h)=\frac{2}{(1+4h^2)^{3/2}} \end{aligned}$$

(13.5)

both of which are finite. Adding an impurity at 0, with spin operators ${\varvec{\tau }}_0$, the Hamiltonian

$$\begin{aligned} H_\lambda =H_0{- h}(\sigma ^z_0+\tau ^z_0) {- \lambda }\sigma ^z_0\tau ^z_0 \end{aligned}$$

(13.6)

is obtained. Does it exhibit a Kondo effect?

Since ${\varvec{\tau }}_0$ commutes with the ${\varvec{\sigma }}_n$ and, hence, with $H_0$, the average magnetization and susceptibility, $m^{int}(\beta ,h,\lambda )$ and $\chi ^{int}(\beta ,h,\lambda )$, responding to a field h acting only on the site 0, can be expressed in terms of the functions $\zeta (\beta ,h)$ and its derivatives $\zeta '(\beta ,h)$ and $\zeta ''(\beta ,h)$. By using the fact that $\zeta (\beta ,h)$ and $\zeta ''(\beta ,h)$ are even in h, while $\zeta '(\beta ,h)$ is odd, we get:

(13.7)

Since $\chi ^{int}(\beta ,0)$ is even in $\lambda$, it diverges for $\beta \rightarrow \infty$ independently of the sign of $\lambda$, while $\chi (\beta ,0)$ is finite. Hence, the model yields Pauli’s paramagnetism, without a Kondo effect.

Remark

• (1) Finally an analysis essentially identical to the above can be performed to study the model in Eq. (2.1) without impurity (and with or without spin) to check that the magnetic susceptibility to a field h acting only at a single site is finite: the result is the same as that of the XY model above: the single site susceptibility is finite and, up to a factor 2, given by the same formula $\chi (\beta ,0)=\frac{4\sinh \beta }{1+\cosh \beta }$.

• (2) The latter result makes clear both the essential roles for the Kondo effect of the spin and of the noncommutativity of the impurity spin components.

Appendix 8: Some Details on Appendix 7

The definition of $H_h$ has to be supplemented by a boundary condition to give a meaning to ${\varvec{\sigma }}_{L+1}$. If $\sigma ^\pm _n=(\sigma ^x\pm i\sigma ^y_n)/2$ define $\mathcal{N}_{<n}$ as $\sum _{i<n}\sigma ^+_i\sigma ^-_i=\sum _{i<n}\mathcal{N}_i$ and $\mathcal{N}=\mathcal{N}_{\le L}$. Then set as boundary condition

$$\begin{aligned} \sigma _{L+1}^\pm \,{\buildrel def\over =}\,-(-1)^\mathcal{N}\sigma ^\pm _{1} \end{aligned}$$

(14.1)

(parity-antiperiodic b.c.) so that $H_h$ becomes

$$\begin{aligned} H_h&=-h (2\sigma ^+_\nu \sigma ^-_\nu - 1) - \frac{1}{2} \sum _{n=1}^{L-1} \left( \sigma ^+_n(-1)^{\mathcal{N}_n}\sigma ^-_{n+1}+ \sigma ^-_n(-1)^{\mathcal{N}_n}\sigma ^+_{n+1}\right) \\&\quad -\,\frac{1}{2} (\,\sigma ^+_L(-1)^{\mathcal{N}_L}(-\sigma ^-_{1}) +\sigma ^-_L(-1)^{\mathcal{N}_L}(-\sigma ^+_{1})\,). \nonumber \end{aligned}$$

(14.2)

Introducing the Pauli–Jordan transformation

$$\begin{aligned} a^\pm _n=(-1)^{\mathcal{N}_{<n}}\sigma ^\pm _n,\quad a^\pm _{L+1}=-a^\pm _1. \end{aligned}$$

(14.3)

In these variables

$$\begin{aligned} H_h={-h(2 a^+_\nu a^-_\nu -1) -\frac{1}{2}} \sum _{n=1}^{L-1} (a^+_na^-_{n+1}- a^-_n a^+_{n+1}) \end{aligned}$$

(14.4)

Assume $L=$even and let $I\,\,{\buildrel def\over =}\,\,\{q| q= \pm \frac{(2n+1)\pi }{L}, \, n=0,1,\ldots ,\frac{L}{2} -1\}$; then

(14.5)

In diagonal form let $U_{jq}$ be a suitable unitary matrix such that

$$\begin{aligned} H_h=\sum _j\lambda _j\alpha ^+_j\alpha ^-_j,\qquad \mathrm{if}\ \alpha ^+_j= \sum _q U_{jq} A^+_q \end{aligned}$$

(14.6)

Then $\lambda _j$ must satisfy

$$\begin{aligned}&\Big (- \sum _q \cos q A^+_q A^-_q -\frac{2h}{L}\sum _{q,q'} A^+_q A^-_{q'}e^{i(q-q')\nu }\Big ) \sum _{q''} U_{jq''} A^+_{q''}{\vert 0\rangle }\nonumber \\&=\lambda _j \sum _{q''} U_{jq''} A^+_{q''}{\vert 0\rangle } \quad \mathrm{hence}\\&(\lambda _j+\cos q)U_{jq}e^{-iq\nu }= -\frac{2h}{L}\sum _{q''} e^{-iq''\nu } U_{jq''},\nonumber \end{aligned}$$

(14.7)

$\forall q\in I$, where we used the fact that $A^-_p A^+_q {\vert 0\rangle } =\delta _{p,q}{\vert 0\rangle }$. We consider the two cases $\lambda _j\ne -\cos q$ for all $q\in I$ or $\lambda =-\cos q_0$ for some $q_0\in I$.

In the first case:

$$\begin{aligned}&U_{jq}=\frac{e^{iq\nu }}{N(\lambda _j)}\frac{1}{\lambda _j+\cos q}, \quad \mathrm{provided}\nonumber \\&F_L(\lambda _j)\,\,{\buildrel def\over =}\,\, 1+\frac{2h}{L}\sum _q\frac{1}{\lambda _j+\cos q}=0, \end{aligned}$$

(14.8)

where ${N(\lambda _j)}$ is set in such a way that U is unitary, or, in the second case,

$$\begin{aligned} \lambda _j=-\cos q_0,\quad U_{jq}=\frac{e^{iq\nu }}{\sqrt{2}}(\delta _{q,q_0}-\delta _{q,-q_0}),\ \ \mathrm{so\ that} \sum _{q''} e^{-iq''\nu }U_{jq''}=0. \end{aligned}$$

(14.9)

Since $-\cos q$ takes $\frac{1}{2}L$ values and the equation $F_L(\lambda )=0$ has $\frac{L}{2}$ solutions, the spectrum of $H_h$ is completely determined and given by the $2^L$ eigenvalues

$$\begin{aligned} \lambda (\mathbf{n})=\sum _j n_j \lambda _j,\quad \mathbf{n}=(n_1,\ldots ,n_L), \ n_j=0,1 \end{aligned}$$

(14.10)

and the partition function is

$$\begin{aligned} \log Z_L(\beta ,h)= & {} \sum _{q>0} \log (1+ e^{\beta \cos q}) \,+\,\sum _{j} \log (1+{ e^{-\beta \lambda _j}} )\nonumber \\= & {} \frac{1}{2}\log Z^0_L(\beta )+\sum _{j\in I} \log (1+{ e^{-\beta \lambda _j}}). \end{aligned}$$

(14.11)

On the other hand, since the function $F'_L(z)/F_L(z)$ has L / 2 poles with residue $+1$ (those corresponding to the zeros of $F_L(z)$) and L / 2 poles with residue $-1$ (those corresponding to the poles of $F_L(z)$), the contour integral in the r.h.s. of Eq. (13.3) is equal to

$$\begin{aligned}&\sum _j \log (1+e^{-\beta \lambda _j}) - \sum _{q>0} \log (1+e^{\beta \cos q})\nonumber \\&\quad =\sum _j \log (1+e^{-\beta \lambda _j}) - \frac{1}{2}\log Z^0_L(\beta )= \log Z_L(\beta ,h) -\log Z^0_L(\beta ). \end{aligned}$$

(14.12)

Appendix 9: meankondo: A Computer Program to Compute Flow Equations

The computation of the flow equation Eq. (10.1) is quite long, but elementary, which makes it ideally suited for a computer. We therefore attach a program, called meankondo and written by I.Jauslin, used to carry it out (the computation has been checked independently by the other authors). One interesting feature of meankondo is that it has been designed in a model-agnostic way, that is, unlike its name might indicate, it is not specific to the Kondo model and can be used to compute and manipulate flow equations for a wide variety of fermionic hierarchical models. It may therefore be useful to anyone studying such models, so we have thoroughly documented its features and released the source code under an Apache 2.0 license. See http://ian.jauslin.org/software/meankondo for details.